C2H2(g) + 2 H2(g)--> C2H6(g)
Substance So (J/mol∙K) ∆Hºf (kJ/mol)
C2H2(g 200.9 226.7
H2(g) 130.7 0
C2H6(g) -- -84.7
Bond Bond Energy (kJ/mol)
C-C 347
C=C 611
C-H 414
H-H 436
If the value of the standard entropy change, ∆Sº for the reaction is -232.7 joules per mole∙Kelvin, calculate the standard molar entropy, Sº, of C2H6
gas.
Calculate the value of the standard free-energy change, ∆Gº, for the reaction. What does the sign of ∆Gº indicate about the reaction above?
Calculate the value of the equilibrium constant for the reaction at 298 K.
Calculate the value of the C C (triple bond) bond energy in C2H2 in kJ/mole.
The standard molar entropy of C2H6 gas is 205.3 J/mol∙K.
The standard free-energy change, ∆Gº, for the reaction is -20.3 kJ/mol. The sign of ∆Gº indicates that the reaction is exothermic.
The equilibrium constant for the reaction at 298 K is 0.0014.
The C C (triple bond) bond energy in C2H2 is 611 kJ/mol.
To calculate the standard molar entropy (Sº) of C2H6 gas, we can use the equation:
∆Sº = ∑nSº(products) - ∑mSº(reactants)
Where n is the stoichiometric coefficient of the product, and m is the stoichiometric coefficient of the reactant.
Using the given data, we can calculate the standard molar entropy of C2H6 gas.
∆Sº = Sº(C2H6) - (2 * Sº(H2)) - Sº(C2H2)
Plugging in the given values:
-232.7 = Sº(C2H6) - (2 * 130.7) - 200.9
Simplifying:
-232.7 = Sº(C2H6) - 261.4 - 200.9
Sº(C2H6) = -232.7 + 261.4 + 200.9
Sº(C2H6) = 229.6 J/mol∙K
Therefore, the standard molar entropy of C2H6 gas is 229.6 J/mol∙K.
To calculate the standard free-energy change (∆Gº) for the reaction, we can use the equation:
∆Gº = ∆Hº - T∆Sº
Where ∆Hº is the standard enthalpy change, T is the temperature in Kelvin, and ∆Sº is the standard entropy change.
Using the given data:
∆Hº = -84.7 kJ/mol
∆Sº = -232.7 J/mol∙K
T = 298 K
Converting ∆Sº to kJ/mol∙K:
∆Sº = -232.7 / 1000 = -0.2327 kJ/mol∙K
∆Gº = -84.7 - (298 * -0.2327)
Simplifying:
∆Gº = -84.7 + 69.3931
∆Gº = -15.3 kJ/mol
The negative sign indicates an exothermic reaction, meaning that the reaction releases energy.
To calculate the value of the equilibrium constant (K) for the reaction at 298 K, we can use the equation:
∆Gº = -RTln(K)
Where R is the gas constant (8.314 J/mol∙K), T is the temperature in Kelvin, and K is the equilibrium constant.
Using the given data:
∆Gº = -15.3 kJ/mol
R = 8.314 J/mol∙K
T = 298 K
Converting ∆Gº to J/mol:
∆Gº = -15.3 * 1000 = -15300 J/mol
-15300 = -8.314 * 298 * ln(K)
Simplifying:
-15300 = -2469.972 * ln(K)
ln(K) = -15300 / -2469.972
ln(K) = 6.201
K = e^(6.201)
K ≈ 501.54
Therefore, the value of the equilibrium constant for the reaction at 298 K is approximately 501.54.
To calculate the value of the C-C (triple bond) bond energy in C2H2, we need to consider the bond energy of a C-C single bond and a C=C double bond.
The bond energy of a C-C single bond is given as 347 kJ/mol.
The bond energy of a C=C double bond can be calculated using the bond energies of C-C single bond and C-H single bond.
C=C bond energy = 2 * (C-C bond energy) + 4 * (C-H bond energy)
C=C bond energy = 2 * 347 + 4 * 414
C=C bond energy = 694 + 1656
C=C bond energy = 2350 kJ/mol
Therefore, the value of the C-C (triple bond) bond energy in C2H2 is approximately 2350 kJ/mol.
To calculate the standard molar entropy, Sº, of C2H6 gas, we can use the standard entropy change formula:
ΔSº = ΣnSº(products) - ΣmSº(reactants)
In this case, since the reaction only involves one product and two reactants, we have:
ΔSº = (1 * Sº(C2H6)) - (2 * Sº(H2) + 1 * Sº(C2H2))
Given that ΔSº = -232.7 J/(mol⋅K), we can rearrange the equation to solve for Sº(C2H6):
-232.7 = Sº(C2H6) - (2 * 130.7 + 1 * 200.9)
Simplifying the equation further, we get:
-232.7 = Sº(C2H6) - 462.3
Sº(C2H6) = -232.7 + 462.3
Sº(C2H6) = 229.6 J/(mol⋅K)
Therefore, the standard molar entropy, Sº, of C2H6 gas is 229.6 J/(mol⋅K).
To calculate the standard free-energy change, ΔGº, for the reaction, we can use the formula:
ΔGº = ΔHº - TΔSº
Given that ΔHº = -84.7 kJ/mol and ΔSº = -232.7 J/(mol⋅K), and converting the temperature to Kelvin (298 K), we have:
ΔGº = -84.7 - (298 * (-232.7 / 1000))
Solving this equation gives us the value of ΔGº in kJ/mol.
The sign of ΔGº indicates the spontaneity of the reaction. If ΔGº is negative, the reaction is spontaneous and thermodynamically favorable. If ΔGº is positive, the reaction is non-spontaneous and thermodynamically unfavorable. If ΔGº is zero, the reaction is at equilibrium.
To calculate the equilibrium constant (K) for the reaction at 298 K, we can use the equation:
ΔGº = -RT ln(K)
R is the gas constant (8.314 J/(mol⋅K)) and T is the temperature in Kelvin.
Given that ΔGº is the value we calculated previously and the temperature is 298 K, we can rearrange the equation to solve for ln(K) and then calculate K:
ln(K) = -ΔGº / (RT)
K = e^(-ΔGº / (RT))
Calculating K using the equation above will give us the value for the equilibrium constant at 298 K.
Finally, to calculate the C-C (triple bond) bond energy in C2H2, we need to subtract the bond energies of the reactants from the bond energies of the products. The bond energy of a molecule is the amount of energy required to break one mole of bonds in the molecule.
In this case, the C-C bond energy in C2H2 can be calculated as follows:
C-C (triple bond) bond energy = Σ(Bond energies of products) - Σ(Bond energies of reactants)
Since C2H6 is the product and C2H2 and H2 are the reactants, we have:
C-C (triple bond) bond energy = (2 * C-C bond energy) - (1 * C=C bond energy + 2 * C-H bond energy + 1 * H-H bond energy)
Substituting the given values for the bond energies, we can calculate the C-C (triple bond) bond energy in C2H2 in kJ/mol.